Neutrino Physics - Lecture 2 Steve Elliott LANL Staff Member UNM Adjunct Professor 505-665-0068,...

Neutrino Physics - Lecture 2

Steve Elliott

LANL Staff Member

UNM Adjunct Professor

505-665-0068, [email protected]

Spring 2007 Steve Elliott, UNM Seminar Series

2

Lecture 2 Outline

• Neutrino detection• Sources of neutrinos

• Neutrino Mixing

• Discussion


3

Neutrino detection

Targets

• H2O

• D2O

• Scintillator

• Ga

• Cl

• Emulsion

• Ice

• Iron

• Rock

ES on e-: x + e--> x + e-

CC on Nucleus: l + A-> A’+ l

NC on Nucleus: x + A-> A’+ x


4

Cross sections

• 10,000 light years of Pb to stop half of solar neutrinos (few MeV e)

• Beta decay provides estimate of strength

€

n → p + e− + ν e

Γ =GF

2

2π 3

mc2

hMif

2f Z, E( )

or : const.

Mif2 = fτ

€

e + p → n + e+

σ 0 =2π 2h3

me5c7 fτ

pe Ee

= 0.0952Ee pe

1MeV2

⎛ ⎝ ⎜

⎞ ⎠ ⎟×10−42 cm2

Neutron beta decay Anti-neutrino absorption


5

Cross Sections

The small size of these cross sections is what led early researchers to believe they had postulated an undetectable particle.


6

Hard experiments

• Rates are very low– Big detectors

• Background difficulties– Signal may not be very distinct– Other more common processes can

mimic signal– Rare variations of common phenomena…


7

Sources of neutrinos

Big BangRadioactive decaysStarsSupernovasCosmic raysReactorsAccelerators


8

Big Bang

• Relic neutrinos contribute at least as much mass to the Universe as all the stars.

• There are as many leftover neutrinos as photons.– N ~420/cc

• Photon energy: 2.728 K• Neutrino energy: 2 K

– There are no viable ideas for detecting such low energy neutrinos.

– But they might have detectable effects for large scale structure

– Note that neutrinos are studied via their particle nature– The microwave background was discovered by the wave

nature of photons.


9

Radioactive Decays

• MCi sources have been made• Mostly for use by solar neutrino radiochemical

experiments for efficiency measurements.• Proposals for other neutrino property

measurements• Electron capture isotopes provide a monoenergetic

neutrino.

51Cr37Ar


10

Stars (our Sun)

FeaturesProduce only e through fusion reactionsVery long baseline, e disappearance, x appearanceLow energy, spectral shape well knownL/E is large so sensitive to small m2

Large FluxMatter enhancement

DataRates from several experimentsEnergy dependenceDay vs. NightSeasonal


11

SupernovasFeatures

~ Very long baseline~ 's and 's~ Complicated and poorly understood source~ Target cross sections not all well understood

Data~ Not a common phenomenon

once ~30 years in our galaxy~ SN1987A provided little physics data~ SN1987A did give hope for the future

10

16

10

14

10

12

10

10

10

8

10

6

10

4

10

2

10

0

(/Flux cm

2

)sec MeV

10

6

10

4

10

2

10

0

( )Energy MeV

supernova

10 @ kpc

My personal prediction is that neutrinos will teach us a lot about supernovae, but the inverse will be much harder.


12

Supernovas

By using various targets with different energy- and flavor-dependent cross sections, one may be able tode-convolute the various fluxes.

Its difficult to get a dedicated supernova neutrino experiment funded.

10

16

10

14

10

12

10

10

10

8

10

6

10

4

10

2

10

0

(/Flux cm

2

)sec MeV

10

6

10

4

10

2

10

0

( )Energy MeV

supernova

10 @ kpc

Some Estimated Rates (Burrows, Klein, Gandhi PR D45, 3361 (1992) Expt.

e p → e

+

n NC on deuterium CC on oxygen

K -II 355 5.5 MACRO 219 Super-K 5310 81.5 SNO 331 272 4


13

Cosmic Rays

atmosphere

Detector

Primary

Cosmic Ray

~20 km

~10000 km

€

Expect Rμ / e =ν μ + ν μ

ν e + ν e≈ 2

Meas.Rμ / edata

Rμ / e MC

≈ 0.6 − 0.7

10

15

10

10

10

5

10

0

10

-5

10

-10

10

-15

(/Flux cm

2

)sec MeV

10

6

10

4

10

2

10

0

( )Energy MeV

supernova

10 @ kpc

π

+

DAR

LSND atmos

CERN SPS

CHORUS


14

Reactors

FeaturesComplicated but well-understood source.Low energyShort, medium, long baselinesDisappearance experiments

DataSeveral at short baselines; 10-250 mCHOOZ/Palo Verde at ~1 kmKamLAND at ~250 km

10

15

10

10

10

5

10

0

10

-5

10

-10

10

-15

Anti-

e

(/Flux cm

2

)sec MeV

10

6

10

4

10

2

10

0

( )Energy MeV

Reactors

CHOOZ

supernova

10 @ kpc

atmos


15

Accelerators

FeaturesUsually appearanceVarious baselines and wide energy rangeControlled experimental conditions

DataOscillation limits for many speciesLots of experimental results

10

15

10

10

10

5

10

0

10

-5

10

-10

10

-15

(/Flux cm

2

)sec MeV

10

6

10

4

10

2

10

0

( )Energy MeV

supernova

10 @ kpc

π

+

DAR

LSND atmos

CERN SPS

CHORUS


16

A reminder of the questions

• Are neutrinos Majorana or Dirac?• What is the absolute mass scale?• How small is 13?• How maximal is 23?• Is there CP violation in the neutrino

sector?• Is the mass hierarchy inverted or normal?• Is the LSND evidence for oscillation true?

Are there sterile neutrinos?


17

Present Values (from oscillation expts.)

€

m122 = (8.0 ± 0.3)10−5 eV 2 Solar/Reactor

δm232 = (2.5± 0.2)10−3 eV 2 Atmospheric

tan2 θ12 = 0.45± 0.05 Solar/Reactor

sin2 θ 23 = 1.02 ± 0.04 Atmospheric

sin2 θ13 ≤ 0.05 Reactor

hep-ph/0606054

Neutrino Oscillations

Or how we know most of what we know


19

Outline

• Two-flavor vacuum oscillations

• Two-flavor matter oscillations

• Three-flavor oscillations– The general formalism

– The “rotation” matrices


20

Consider Two Mass States

1 corresponding to m1

2 corresponding to m2

Think of as a Vector

€

Ψ =1(t)

ψ2 (t)

⎛

⎝ ⎜

⎞

⎠ ⎟=

e−iE1t ψ1

e−iE2t ψ2

⎛

⎝ ⎜

⎞

⎠ ⎟

€

E1 = p12 + m1

2 ≈ p1 +m1

2

2 p1


21

Ψ is a solution of H

€

i∂∂t

Ψ = HΨ

€

H=E1 0

0 E2

⎛

⎝ ⎜

⎞

⎠ ⎟

€

i∂ /∂t ψ1(t)

i∂ /∂t ψ2 (t)

⎛

⎝ ⎜

⎞

⎠ ⎟=

E1 0

0 E2

⎛

⎝ ⎜

⎞

⎠ ⎟ψ1(t)

ψ2 (t)

⎛

⎝ ⎜

⎞

⎠ ⎟


22

The Neutrinos

Consider the weak eigenstates e, .These are not the mass eigenstates, 1, 2.The mass eigenstates are propagated via H.

€

e (t)

ν μ (t)

⎛

⎝ ⎜

⎞

⎠ ⎟=

cosθ sinθ

−sinθ cosθ

⎛

⎝ ⎜

⎞

⎠ ⎟ν 1(t)

ν 2 (t)

⎛

⎝ ⎜

⎞

⎠ ⎟

The Mixing Matrix: U


23

Mixing

Weak eigenstates are a linear superposition of mass eigenstates.

€

α =Uν i


24

In Vacuum, no potential in H

€

−i∂∂t

ν i = Hν i

−i∂∂t

U−1ν α = HU−1ν α

−i∂∂t

ν α = UHU−1ν α

Denote c = cos s = sin


25

UHU-1

€

UHU−1 =c s

−s c

⎛

⎝ ⎜

⎞

⎠ ⎟E1 0

0 E2

⎛

⎝ ⎜

⎞

⎠ ⎟c −s

s c

⎛

⎝ ⎜

⎞

⎠ ⎟

€

= E1C2 + E2S2( )

1 0

0 1

⎛

⎝ ⎜

⎞

⎠ ⎟+

0 sc(E2 − E1 )

sc(E2 − E1 ) (c2 − s2 )(E2 − E1 )

⎛

⎝ ⎜

⎞

⎠ ⎟


26

The energy difference (and Trig.)

€

E2 − E1 = p +m2

2

2 p− p −

m12

2 p

E2 − E1 =m2

2 − m12

2 p≡

δm2

2 p≈

δm2

2E

€

2sc ≡ sin2θ

c2 − s2 ≡ cos2θ


27

UHU-1 becomes

€

= E1C2 + E2S2( )

1 0

0 1

⎛

⎝ ⎜

⎞

⎠ ⎟+

δm2

4E

0 sin2θ

sin2θ 2cos2θ

⎛

⎝ ⎜

⎞

⎠ ⎟

The algebra is going to get involved, so lets defineA, B, and D such that:

€

UHU−1 =A B

B A+ D

⎛

⎝ ⎜

⎞

⎠ ⎟


28

The Diff Eq

€

∂α∂t

= iA B

B A+ D

⎛

⎝ ⎜

⎞

⎠ ⎟ν α

A solution to this equation should have the form:

€

α =e

ν μ

⎛

⎝ ⎜

⎞

⎠ ⎟=

ye

yμ

⎛

⎝ ⎜

⎞

⎠ ⎟eirt


29

Insert proposed solution

€

rye

yμ

⎛

⎝ ⎜

⎞

⎠ ⎟eirt =

A B

B A+ D

⎛

⎝ ⎜

⎞

⎠ ⎟ye

yμ

⎛

⎝ ⎜

⎞

⎠ ⎟eirt

or

A− r B

B A+ D − r

⎛

⎝ ⎜

⎞

⎠ ⎟ye

yμ

⎛

⎝ ⎜

⎞

⎠ ⎟= 0


30

Two Equations

€

(A− r)ye + Byμ = 0

Bye + (A+ D − r)yμ = 0

(A− r)(A+ D − r)− B2 = 0

⇒ r =D + 2A± D + 4B2

2


31

r+ solution

€

A− r+ B

B A+ D − r+

⎛

⎝ ⎜

⎞

⎠ ⎟ye

yμ

⎛

⎝ ⎜

⎞

⎠ ⎟= 0

ye =−sinθcosθ

yμ

r- solution

€

ye =cosθsinθ

yμ


32

α is a superposition of these 2 solutions

€

α =C1sinθ

cosθ

⎛

⎝ ⎜

⎞

⎠ ⎟e

ir+t + C2cosθ

−sinθ

⎛

⎝ ⎜

⎞

⎠ ⎟e

−ir−t

r± =12

(D + 2A)±δm2

4E

(D+2A) is a constant so we sweep it into a redefinition of the C’s.


33

The solutions

€

α =C1sinθ

cosθ

⎛

⎝ ⎜

⎞

⎠ ⎟e

iδm2

4Et

+ C2cosθ

−sinθ

⎛

⎝ ⎜

⎞

⎠ ⎟e

−iδm2

4Et

To determine the C’s, use <α|α>=1 and assume that at t=0, we have all e.

€

α (t = 0) =1

0

⎛

⎝ ⎜

⎞

⎠ ⎟=

C1 sinθ + C2 cosθ

C1 cosθ − C2 sinθ

⎛

⎝ ⎜

⎞

⎠ ⎟

€

⇒ C1 = sinθ , C2 = cosθ


34

The time dependent solution

€

α =sin2 θ

sinθ cosθ

⎛

⎝ ⎜

⎞

⎠ ⎟e

iδm2

4Et+

cos2 θ

−sinθ cosθ

⎛

⎝ ⎜

⎞

⎠ ⎟e

−iδm2

4Et

What is the probability of finding all at time t?

€

α (t) = prob. amp.

= sinθ cosθeiδm2

4Et− sinθ cosθe−

iδm2

4Et

= 2sinθ cosθ (12

eiK − e−iK[ ] = sin2θ sin

δm2

4Et

⎛

⎝ ⎜

⎞

⎠ ⎟


35

Transition probability

€

α (t)2

= probability

= sin2 2θ sin2 δm2

4Et

⎛

⎝ ⎜

⎞

⎠ ⎟

€

define δm2

4Et ≡

πRL

for h = c = 1, R ≈ t

L =4πE

∂m2 ≡ oscillation length


36

The Answer

€

P(ν e → ν μ ) = sin2 2θ sin2 1.27δm2 (eV2 )E(MeV)

R(meters) ⎛

⎝ ⎜

⎞

⎠ ⎟

Complete mixing: large sin2 and long R/L would result in an “average”: that is P=1/2.

Neutrino Physics - Lecture 2 Steve Elliott LANL Staff Member UNM Adjunct Professor 505-665-0068,...

Documents

Transcript of Neutrino Physics - Lecture 2 Steve Elliott LANL Staff Member UNM Adjunct Professor 505-665-0068,...